User:Struggle Town Mayor/MATH5515

Solutions from Tutes

Special Relativity (1-23)

 * 1) Two light signals sent $d$ time apart. Draw picture, $kd$ is clear.
 * 2) tutes
 * 3) asdf
 * 4) sub $\beta$; find $\frac{dv}{d\beta}$ to get $\beta^3 vdv = \cdots = \frac{\beta v}{\sqrt{\beta^2-1}} = c$; $\frac{d}{dv}(\beta v) = \beta + v\frac{d\beta}{dv} = \beta + v\frac{\beta^3 v}{c^2} = \beta^3$.
 * 5) equation 1.6 from notes is straight from definition. boost to $I'$ with rapidity $\lambda$ and then to $I' '$ with rapidity $\mu$ gives $t' ' = t\cosh(\lambda + \mu) - x\sinh(\lambda + \mu)$
 * 6) (a) we get $\beta = \frac{5}{3}$, so $v = \frac{4}{5}$ and $x' = -5v = -4$. (b) $s(E'-E,E'-E) = -3 < 0 \implies$ spacelike separated. (b) (i) $E$ and $E'$ simultaneous if $1-2v = 0 \iff v=\frac{1}{2}$. (ii) solving \[-1 = \frac{1}{\sqrt{1-v^2}}(1-2v)\] gives $v = 0$ (for $E$ occurring $1$ unit of time before $E'$) and $v = \frac{4}{5}$ ($E$ occurring $1$ unit of time after $E'$).
 * 7) need length to contract so that in magician's frame $l \leq 1.5$. $l = 1.5$ at $\beta = \frac{4}{3}$, find $v$. in assistant's frame, take $x = 0$ to be midpoint between two blades, suppose blades fall at $t=0$. we have \[ \left( \begin{matrix} \beta & -\beta v & 0 & 0 \\ -\beta v & \beta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right) \left( \begin{matrix} 0 \\ \pm \frac{3}{4} \\ 0 \\ 0 \end{matrix} \right) = \left( \begin{matrix} \mp \frac{\sqrt{7}}{4} \\ \pm 1 \\ 0 \\ 0 \end{matrix} \right),\] i.e. the positive coordinate blade drops first.
 * 8) I still don't like this one :( http://en.wikipedia.org/wiki/Ladder_paradox
 * 9) asdf
 * 10) $(t_1,x_1)$ and $(t_2,x_2)$ are Minkowski orthogonal iff $t_1t_2 = x_1x_2$, so the direction of their slopes are inverse to each other.
 * 11) since a timelike 2-space has a timelike vector in its basis, Lemma 1.3 shows the orth. complement is spacelike. for "vice versa", any direction orthogonal to a spacelike 2-space is, wlog, orthogonal to $(0,0,\alpha,\beta)$. 2-dimensional implies there is a basis with a timelike vector. for a null 2-space, Lemma 1.4 says a vector orthogonal to a null vector is null or spacelike. but if we have a basis of two spacelike vectors in the complement, then the original space is timelike, so the complement must be spanned by a parallel null vector and a spacelike vector. (two parallel null vectors would only give 1-d space).
 * 12) too much for my calculator wat do
 * 13) asdf
 * 14) length contract the rod with $0.8c$, and find relative velocity to get time (keep everything in $I$ frame)
 * 15) not sure what the first part means - nothing changes in the $x$/$x'$ direction, nothing really for $y$/$y'$ too, except to scale by $\beta^{-1}$ for time dilation? might have the wrong idea here.
 * 16) tutes
 * 17) tutes
 * 18) (a) verifying that \[\beta\left(\frac{1+t}{\sqrt{2+2t+t^2}}\right)=\sqrt{2+2t+t^2}.\] (b) For mass, $s(P,P) = 16$ so that $m=4$. Do $s(P,U)$ to find energy
 * 19) rearrange \[ E = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}\] to get \[v = c\sqrt{1-\frac{m^2c^4}{E^2}}.\] take first order taylor expansion, yay.
 * 20) Have \begin{align}m_1\beta_1 + m_2\beta_2 &= m_3\beta_3 \\ m_1\beta_1u_1 + m_2\beta_2u_2 &= m_3\beta_3u_3\end{align} then you just chug out one of the more disgusting sim. equs to get it. (More detail if you need): Start by trying to show the second one by considering the first component of the conservation of momentum equation (so you can just cancel c's rather than not being able to cancel different velocity components.) Then you sub in $\beta_3 = (1 - \frac{u_{3}^2}{c^2})^{-1/2}$ and play around to get the result.
 * 21) (Needs checking, I'm pretty sure I got this wrong cause I had it retardedly messy.) \[ v = \frac{h \nu c}{mc^2 + h\nu} \qquad m' = \sqrt{m^2 + \frac{2mh\nu}{c^2}}\]
 * 22) Suppose a photon travels to the origin in the $x$-$y$ plane. Suppose it starts at $P = (0,\cos(\alpha),\sin(\alpha))$ and gets there at $Q = (1,0,0)$. In a frame moving in the $-x$ direction (keeps the angle going the right way) at speed $v$, we have \[ P' = v\beta\cos(\alpha),\beta\cos(\alpha),\sin(\alpha)) \qquad Q' = (\beta,v\beta,0). \] To get the angle in this frame, we see that $\cos(\alpha') = \displaystyle \frac{x(P') - x(Q')}{t(Q')-t(P')}$ (reversing $P'$ and $Q'$ in numerator keeps things positive, and difference in $t$ gives how far it travels in total because it's going light speed), and plugging stuff in gives \[ \cos(\alpha') = \frac{\cos(\alpha)-v}{1-v\cos(\alpha)}. \]
 * 23) from tute, it's a pain.

Differential Geometry (24-43)
screw ordered numbering; this junk is too hard.

  first, in Cartesians on $\mathbb E^3$, \[ X^a_{\text{ };a} = X^a_{\ ,a} + \Gamma^a_{ac} X^c = X^a_{\ ,a}\] which is the normal divergence. in general \[ X^a_{\text{ } ;a} = X^a_{\ ,a} + \Gamma^a_{ac} X^c = X^a_{\ ,a} + \frac{1}{2}\ln(|g|)_{,c}X^c = X^a_{\ ,a} + \frac{1}{\sqrt{|g|}}(\sqrt{|g|})_{,c}X^c = \frac{1}{\sqrt{|g|}}(\sqrt{|g|}X^a)_{,a}\]

 Let $X^a = (X^0, X^1)$ Write $g_{ab}$ and $g^{ab}$ and use them to find $X_a = (y X^0,x^2 X^1)$. Find Christofells - only nonzero symbols are $\Gamma_{10}^0 = \Gamma_{01}^0 = \frac{1}{2y}$, $\Gamma _{00}^1 = \frac{-1}{2y^2}$, $\Gamma_{11}^1 = \frac{1}{y}$.

For $X^{a}$ to be a Killing vector, we need $X_{a;b} + X_{b;a} = 0$. So this will give $X_{0;0} = 0, X_{1;0} + X_{0;1} = 0, X_{1;1}=0$.

From these we get the following three DEs:

1. $y \frac{\partial X^1}{\partial x} + \frac{\partial X^0}{\partial y} = 0$

2. $2y \frac{\partial X^0}{\partial x} + X^1 = 0$

3. $yX^1 + y^2 \frac{\partial X^1}{\partial y} = 0.$

We can solve the DE in (1) (If we remember how to...) to get $X^1 = \frac{f(x)}{y}$ sub into (1) to get \[ f'(x) + \frac{\partial X^0}{\partial y} = 0 \] \[ \frac{\partial X^0}{\partial y} = -f'(x) \] \[ X^0 = -yf'(x) + g(x) \] $$ \frac{\partial X^0}{\partial x} = -yf''(x) + g'(x) $$

Now, our equation (2) had a $\frac{\partial X^0}{\partial x}$ and we sub this in to get

\[ -2y^2 f''(x) + g'(x) +X^1 = 0 \] So, (using $X^1 = f(x)/y$)

$f(x) = 2y^3 f''(x) - y g'(x)$ So, for this to be a function of x only we need $f'', g' =0$ so that f is identically 0. This gives that the only killing vector is the obvious $\partial_{x}$. Note that this is obvious because all the coefficients in the metric are independent of $x$, see proof at bottom of page. 

Spacetimes (44-55)
  Sub in $l_a = (t_a+z_a)/\sqrt{2}$ and $n_a$ into $l_an_b+l_bn_a$ and you get $t_at_b-z_az_b$. Same for the $m_a, \overline{m}_b$ (p3, (3.1)) to get $x_ax_b+y_ay_b$.



(a)We have \begin{align*}N^a_bN^b_c&=(\ell^ax_b-x^a\ell_b)(\ell^bx_c-x^b\ell_c)\\&=\ell^ax_b\ell^bx_c-\ell^ax_bx^b\ell_c-x^a\ell_b\ell^bx_c+x^a\ell_bx^b\ell_c\\ &=\ell^a\ell_c.\end{align*}

Also $N^3=N^a_bN^b_cN^c_d=\ell^a\ell_c(\ell^cx_d-x^c\ell_d)=\ell^a\ell_c\ell^cx_d-\ell^a\ell_cx^c\ell_d=0$ so clearly $N^3=0$ for all $k\geq3$.

$\newcommand{\l}{\ell}$ (b) $\exp(tN)^a_b\l^b=(1+tN+\frac{t^2}{2}N^2)\l^b=\l^b+t\l^a\l_b\l^b\l_a+\frac{t^2}{2}(\l^ax_b-x^a\l_b)\l^b=\l^b$ Hence is a null rotation

(c) We have $T^a_b=t^ax_b-x^at_b$ and $X^a_b=(x^ay_b-y^ax_b)$ by definition(I lowered the index without using the metric...surely thats legit...? $\leftarrow$ yeh it works out).

Now, notice that \begin{align*}T^a_b t^b&=-x^a\\T^a_bx^b&=-t^a\end{align*} and so \begin{align*}\exp(\lambda T)^a_bt^b&=t^a-\lambda x^a+\frac{\lambda^2}{2!}t^a-\frac{\lambda^3}{3!}x^a+...\\=&(1+\frac{\lambda^2}{2!}+\frac{\lambda^4}{4!}+...)t^a-(\lambda+\frac{\lambda^3}{3!}+\frac{\lambda^5}{5!})x^a\\&=\cosh(\lambda)t^a-\sinh(\lambda)x^a.\end{align*} Similarly $\exp(\lambda T)^a_bx^b=-\sinh(\lambda)t^a+\cosh(\lambda)x^a$

Similarly we have \begin{align*}X^a_bx^b&=y^a\\X^a_by^b&=-x^a\end{align*} so \begin{align*}\exp(\theta X)^a_bx^b&=x^a+\theta y^a-\frac{\theta^2}{2!}x^a-...\\&=(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-...)x^a+(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-...)y^a\\&=\cos(\theta)x^a+\sin(\theta)y^a.\end{align*} Also we have $\exp(\theta X)^a_by^b=-\sin(\theta)x^a+\cos(\theta)y^a$.

 tute week 8

 i) $v = 0.96c \Rightarrow t = 1/0.96c =3.47 \times 10^{-10}$s. $r = 0.5gt^2$ ($\leftarrow$ think that's legit (?)) gives $r = 5.9 \times 10^{-19}$ smaller than nucleii so no good ii) sub in $r = 10^{-9}$ to get $t = 1.43 \times 10^{-5}$ and $t\times 0.96c = 4114.2$. (?)

 oh just realised j steele has answers on blackboard :| incl this one

 j Steele blackboard



Solutions of the Field Equations (56-60)
  As in the lecture notes p.5 look at $L_{X_{2}}g_{23}$. Since $g_{23}=0$ we consider the stuff with $X_{2}^2 = sin(\theta)$ and $X_{2}^3 = cos(\theta)cot(\phi)$ and since the Lie Derivative is zero we get what we came for.

(a)Null is immediate because there's no $dv^2$ term in the metric. Killing because the components of the metric are $v$-independent (see end of page). Geodesic... [???] note for the lie derivative since no terms in the metric depend on $v$ we have that $g_{ab,1} = 0$ all the time. Also, $X_{1,b}^c$ is always going to be zero since only have $X_{1}^1 =1$ which will disappear.

(b) We have \[ g_{ab} = \left(\begin{smallmatrix} 2H&1&0&0\\ 1&0&0&0 \\ 0&0&-1&0 \\ 0&0&0&-1 \end{smallmatrix} \right) \] and \[ g^{ab} = \left(\begin{smallmatrix} 0&1&0&0\\ 1&-2H&0&0 \\ 0&0&-1&0 \\ 0&0&0&-1 \end{smallmatrix} \right) \] Signature is found if we diagonalize the matrix like so (see J) http://www.wolframalpha.com/input/?i=diagonalize+{{2H%28x%2Cy%2Cu%29%2C1%2C0%2C0}%2C{1%2C0%2C0%2C0}%2C{0%2C0%2C-1%2C0}%2C{0%2C0%2C0%2C-1}} and we count the number of positive and negative signs. Three negative and one positive gives signature of -2. Nonzero Christoffel's are $\Gamma_{00}^1 = H_{,u}, \Gamma_{20}^1 = H_{,x}, \Gamma_{30}^1 = H_{,y}, \Gamma_{00}^2 = H_{,x}, \Gamma_{00}^3 = H_{,y}$. None of these have a zero [do you mean 1? - Jason] in the subscript so the covariant derivative will end up being zero(i.e. cov. constant). A curve with $u,x,y$ constant is geodesic, as the tangent vector has form $X^a = f\partial_v$, $f$ a function, so that $X^a_{\text{ };b} = f_{,b}\delta^a_v$. This means $X^a_{\text{ };b}X^b$ is parallel to $X^a$, so the curve is geodesic. (To see that last bit, observe that $X^a_{\text{ };b}$ is like a matrix with only the $v$ row non-zero, so the product with $X^b$ will stay parallel to $\partial_v$.)

(c) Two things to keep in mind: (1) The antisymmetry in the last two indices. (2) $\Gamma^0_{**} = 0$. Follows immediately that $R^0_{bcd} = 0$ for all $b,c,d$. The hint shows that $R^a_{bcd} = 0$ if $c$ or $d$ is $1$; what we saw before about no $1$s in the bottom components of Christoffels shows that $R^a_{bcd} = 0$ if $b = 1$ as well, so from now we'll ignore those cases.

Argue by trying to keep components non-zero as much as possible, but we'll find that it often gets impossible. Consider \[ R^1_{bcd} = \partial_c\Gamma^1_{bd}-\partial_d\Gamma^1_{bc}+\Gamma^e_{bd}\Gamma^1_{ce}-\Gamma^e_{bc}\Gamma^1_{ed}. \] Start with the last term, $\Gamma^e_{bc}\Gamma^1_{ed}$. For $\Gamma^1_{ed}$ to be non-zero we need $e$ or $d$ to be zero, and the other to not be $1$. But if $e = 0$, the whole thing vanishes by (2), so then $d = 0$ and $e \neq 1$. But now this requires $b = 0 = c$, so $c = d$ and by (1) the whole thing vanishes. Similar for the second last term. We're left with \[ R^1_{bcd} = \partial_c\Gamma^1_{bd}-\partial_d\Gamma^1_{bc}. \] If $b \neq 0$, then $c = 0 = d$ so by (1) the whole thing vanishes again, so we're limited to $b = 0$. If either $c$ or $d$ is $1$, one term will get knocked off, the other involves differentiating $H$ with respect to $v$, so again it vanishes. If both are not equal to $1$, it ends up in the form $H_{,dc} - H_{,cd} = 0$. So all $R^1_{bcd} = 0$.

Now, for $a = 2$, \[ R^2_{bcd} = \partial_c\Gamma^2_{bd}-\partial_d\Gamma^2_{bc}+\Gamma^e_{bd}\Gamma^2_{ce}-\Gamma^e_{bc}\Gamma^2_{ed}. \] The last two terms are a lot easier to knock off this time, because the second Christoffel in each one forces $e = 0$, which makes the first Christoffel vanish by (2). We're left with \[ R^2_{bcd} = \partial_c\Gamma^2_{bd}-\partial_d\Gamma^2_{bc}. \] We'll need $b = 0$, and the other two must be $0$ and either $2$ or $3$. Exactly the same for $a = 3$. Finally, the only non-zero components we get (up to antisymmetry) are: \begin{align*} R^2_{020} &= H_{,xx}\\ R^2_{030} &= H_{,xy}\\ R^3_{020} &= H_{,xy}\\ R^3_{030} &= H_{,yy}. \end{align*}

(d) We have $R_{ab} = R^c_{acb} = H_{,xx} + H_{,yy} = 0$ if $H$ is harmonic.

(e) $\partial_u$? Really ceebs with the rest of this, I don't know if there's way to do it other than more monkey work.

 (please check all calculations for this question)

(a) Using the redshift formula we get fractional change of wavelength as $5*10^-7$ so that 656nm $\rightarrow$ 656.000328nm. Remember for schwarzschild $\nu \rightarrow 0$ as you go to infinity.

(b) Get factor of 4/3 coming about (1 +1/3) so $656nm \rightarrow 874.66nm$.

(c) 980.752

For the speed/special rel. bit use $c = \nu \lambda$ with $\lambda = 656 \times 10^-9$ to find frequency $\nu$. Then formula 1.17 will give you what you want.

</ol>

Black Holes (61-72)
<ol>

</li> First note equation (5.4) in the notes so that at a minimum of $V(r)$ we have $\ddot{r} = \frac{L^2}{r^3} - \frac{3mL^2}{r^4} - \frac{m \alpha}{r^2} = 0 = -\frac{\partial V}{\partial r}$. Remember $\alpha = 1$ for a material particle. So we can get $L^2 = \frac{mr^2}{r-3m}$. We also have that $\dot{r} =0$ (in a circular orbit so the radius doesn't change) so equation (5.2) in the notes says that $\frac{1}{2} E^2 = V(r) = \frac{1}{2}(1-2m/r)(1+L^2/r^2)$ put in our value of $L^2$ to get (after a little algebra) the result. Minimum radius circular orbit $r=6m$ plug in to get $E(6m) = \sqrt{8}/3$. So energy loss is $1- \sqrt{8}/3$ is about $6 percent$.

</li>For null, just let it be $X^a$ and lower the index and inner product is 0. For the $\kappa = \sigma = 0$ bit look at Q48 and it says there's some Riemann tensor condition that will give it - calculate if you want.

</li> First part since $u = t - logy$ and $v= t + logy$ we find $du = dt - y^{-1} dy$ and $ dv = dt + y^{-1} dy$. The Rindler metric is $ds^2 = dt^2 - y^{-2} dy^2$ but this is just $y du dv$. However, $y = exp(1/2(v-u))$. Now to show $e^{\frac{v}{2}}$ is an affine parameter along outgoing null geodesics. Ch5 p. 11 tells what we need to do. In particular it says for an affine parameter $\lambda$ we have that the quantity $y\frac{dt}{d\lambda}$ is constant, say, $K$. Then integrate to find $\lambda$. $\lambda = (1/K) \int e^{\frac{v-u}{2}} dt$ but for outgoing null geodesics $u$ is a constant so we can factor the $e^{-u/2}$ out and also we have that $dt = 1/2 dv + 0$ ($du = 0)$. So we get $\lambda = (constant)*e^{v/2}$ which is good.

</li> a) (We did this in the last tute but it's not on BB so here here it is) \[ r = \rho(1+2m/ \rho)^2 \] \[ dr = d\rho(1+2m/\rho)^2 + 2\rho(1+2m/ \rho)(-m/2\rho^2)d\rho \\ \\= (1+m/2\rho)(1-m/2\rho)d\rho = 1/4\rho^2 (\rho + 2m)(\rho - 2m)d\rho \] and \[ 1-2m/r = 1- 2m/\rho(1 + m/2\rho)^-2 \\= 1- 2m/\rho((2\rho+m)/2\rho)^-2 \\=((2\rho - m)/(2\rho + m))^2 \] So, \[ (1 - 2m/r)dt^2 = ((2\rho - m)/(2\rho + m))^2dt^2 \] and then \[ (1-2m/\rho)^-1 dr^2 = ((2\rho + m)/(2\rho - m))^2*(1/16\rho^4)(2\rho + m)^2(2\rho - m)^2 d\rho^2 \\ = (1+m/2\rho)^4d\rho^2 \] and $r^2 = \rho^2(1 + m/2\rho)^4$  - Hence, the result follows. \\Now, looking at the metric - we see it's asymptotically flat and regions I and IV are the two asymptotically flat regions in the Kruskal- Szekeres diagram. It could be either of these regions, we show that it's both. We put $r= \rho(1 + m/2\rho)^2$ into Kr.Sz. coordinates. If you look on the fact sheet for Kruskal-Szekeres solution you get that $-(r-2m)e^{-r/2m} = T^2 - X^2$. \[ =-r(1-2m/r)e^{-r/2m} \\= -r((2\rho -m)/(2\rho + m))^2e^{-r/2m} \] which is negative. So we have $|T| < |X|$ which is exactly regions $I$ and $IV$. (According to JS the rest is less interesting but will be put up on BB - lies)

b) (feels a bit dodgy, please check) Proper distance on the $\rho$-axis will be found if we let the $dt^2 = d\theta^2 = d\phi^2 = 0$ (I saw the letting $dt^2 = 0$ in a book and the rest I kind of assumed since we want to be on the $\rho$-axis so that the other coordinates are zero - (as far as I know we haven't done stuff like this before - hopefully someone can correct this if necessary). Then we get a separable DE. </ol>

$g_{ab}$ independent of variable indexed $k$ implies $X^a = \partial_k$ is a Killing vector.
We have \begin{align*} X_{a;b} &= g_{ak,b} - \Gamma^c_{ab}g_{ck}\\ &= g_{ak,b} - \frac{1}{2}g^{cd}(g_{ad,b}+g_{bd,a}-g_{ab,d})g_{ck}\\ &= g_{ak,b} - \frac{1}{2}(g_{ak,b}+g_{bk,a}-g_{ab,k}) \qquad \text{since $g^{cd}g_{ck} = \delta^d_k$}\\ &= \frac{1}{2}(g_{ak,b}-g_{bk,a}) \qquad \text{since $g_{ab,k} = 0$}, \end{align*} which is antisymmetric.

$\newcommand{\zomgmacro}{\Large{ZOMG\,\,\,\,MACRO\,\,\,\,WORKS}} \zomgmacro$

(Possible) Open Questions

 * 1) Magician paradox (tute Q7) - Simultaneity - see http://www.parabola.unsw.edu.au/vol35_no1/node2.html
 * 2) Barn paradox (tute Q8) - see same link as above for JS's explanation.
 * 3) Twin Paradoxes(tute Q13) - see "Resolution of the paradox in special relativity" section in http://en.wikipedia.org/wiki/Twin_paradox
 * 4) Principle of equivalence
 * 5) Mach principle
 * 6) Olbers' paradox (ch6)